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Find an expression for the electric field strength at a distant point situated (i) on the axis and (ii) along the equatorial line of an electric dipole.
OR
Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment $\vec{\text{p}}$ and length 2a. What is the direction of this field?

Vidyadip · Accepted Answer

Consider an electric dipole AB. The charges -q and +q of dipole are situated at A and B respectively as shown in the figure. The separation between the charges is 2a. Electric dipole moment, p = q.2a The direction of dipole moment is from -q to +q.Image

At axial or end-on position: Consider a point P on the axis of dipole at a distance r from mid-point O of electric dipole.

The distance of point P from charge +q at B is BP = r - a.
And distance of point P from charge -q at A is, AP = r + a.
Let $E_1$ and $E_2$ be the electric field strengths at point P due to charges +q and -q respectively.
We know that the direction of electric field due to a point charge is away from positive charge and towards the negative charge. Therefore,
$\text{E}_1=\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}-\text{a})^2}$ (from B to P)
and $\text{E}_2=\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}+\text{a})^2}$ (from P to A)
Clearly the directions of electric field strengths $\vec{\text{E}}_1$ and $\vec{\text{E}}_2$ along the same line but opposite to each other and $E_1 > E_2$ because positive charge is nearer.
$\therefore$ The resultant electric field due to electric dipole has magnitude equal to the difference of $E_1$ and $E_2$ direction from B to P i.e.,
$E = E_1 - E_2$
$=\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}-\text{a})^2}-\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}+\text{a})^2}$
$=\frac{\text{q}}{4\pi\in_0}=\Big[\frac{1}{(\text{r}-\text{a})^2}-\frac{1}{(\text{r}+\text{a})^2}\Big]=\frac{\text{q}}{4\pi\in_0}\bigg[\frac{(\text{r}+\text{a})^2-(\text{r}-\text{a})^2}{(\text{r}^2-\text{a}^2)^2}\bigg]$
$=\frac{\text{q}}{4\pi\in_0}\frac{4\text{ra}}{(\text{r}^2-\text{a}^2)^2}=\frac{1}{4\pi\in_0}\frac{2(\text{q}2\text{a})\text{r}}{(\text{r}^2-\text{a}^2)^2}$
But q.21 = p (electric dipole moment)
$\therefore\ \text{E}=\frac{1}{4\pi\in_0}\frac{2\text{pr}}{(\text{r}^2-\text{a}^2)^2}\dots(\text{i})$
If the dipole is infinitely small and point P is far away from the dipole, then r >> l, therefore equation (i) may be expressed as,
$\text{E}=\frac{1}{4\pi\in_0}\frac{2\text{pr}}{\text{r}^4}$ or $\text{E}=\frac{1}{4\pi\in_0}\frac{2\text{p}}{\text{r}^3}\dots(\text{ii})$
This is the expression for the electric field strength at axial position due to a short electric dipole.

At a point of equatorial line: Consider a point P on broad side on the position of dipole formed of charges +q and -q at separation 2a. The distance of point P from mid-point (O) of electric dipole is r. Let $\vec{\text{E}}_1$ and $\vec{\text{E}}_2$ be the electric field strengths due to charges +q and -q of electric dipole.

From fig. $\text{AP}=\text{BP}=\sqrt{\text{r}^2+\text{a}^2}$
$\therefore\ \vec{\text{E}}=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{r}^2+\text{a}^2}$ along B to P
$\vec{\text{E}}_2=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{r}^2+\text{a}^2}$ along P to A
Clearly $\vec{\text{E}}_1$ and $\vec{\text{E}}_2$ are equal in magnitude i.e., $|\vec{\text{E}}_1|=|\vec{\text{E}}_2|$ or $E_1 = E_2$
To find the resultant of $\vec{\text{E}}_1$ and $\vec{\text{E}}_2,$ we reslove them into rectangular components.
Component of $\vec{\text{E}}_1$ Parallel to $\text{AB}=\text{E}_1\cos\theta,$ in the direction to $\vec{\text{BA}}$
Component of $\vec{\text{E}}_1$ perpendicular to $\text{AB}=\text{E}_1\sin\theta,$ along OP
Component of $\vec{\text{E}}_2$ Parallel to $\text{AB}=\text{E}_2\cos\theta,$ in the direction to $\vec{\text{BA}}$
Component of $\vec{\text{E}}_2$ perpendicular to $\text{AB}=\text{E}_2\sin\theta,$ along PO
Clearly, comoponents of $\vec{\text{E}}_1$ and $\vec{\text{E}}_2$ perpendicular to $\text{AB}=\text{E}_1\sin\theta$ and $\text{E}_2\sin\theta$ being equal and opposite cancel each other, while the components of $\vec{\text{E}}_1$ and $\vec{\text{E}}_2$ parallel to $\text{AB}=\text{E}_1\cos\theta$ and $\text{E}_2\cos\theta,$ being in the same direction add up and give the resultant electric field whose direction is parallel to $\vec{\text{BA}}.$
$\therefore$ Resultant electric field at P is $\text{E}=\text{E}_1\cos\theta+\text{E}_2\cos\theta$
But $\text{E}_1=\text{E}_2=\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}^2+\text{a}^2)}$
From the figure, $\cos\theta=\frac{\text{OB}}{\text{PB}}=\frac{\text{l}}{\sqrt{\text{r}^2+\text{a}^2}}=\frac{\text{l}}{(\text{r}^2+\text{a}^2)^{1/2}}$
$\text{E}=2\text{E}_1\cos\theta=2\times\frac{1}{4\pi\in_0}\frac{\text{q}}{(\text{r}^2+\text{a}^2)}.\frac{\text{l}}{(\text{r}^2+\text{a}^2)^{1/2}}$
$=\frac{1}{4\pi\in_0}\frac{2\text{ql}}{(\text{r}^2+\text{a}^2)^{3/2}}$
But q.2l = p = electric dipole moment ...(iii)
$\therefore\ \text{E}=\frac{1}{4\pi\in_0}\frac{\text{p}}{(\text{r}^2+\text{a}^2)^{3/2}}$
If dipole is infinitesimal and point P is far away, we have a << r, so $l^2$ may be neglected as compared to $r^2$ and so equation (iii) gives,
$\text{E}=\frac{1}{4\pi\in_0}\frac{\text{p}}{(\text{r}^2)^{3/2}}=\frac{1}{4\pi\in_0}\frac{\text{p}}{\text{r}^3}$
i.e., electric field strength due to a short dipole at broadside on position.
$\text{E}=\frac{1}{4\pi\in_0}\frac{\text{p}}{\text{r}^3}$ in the direction parallel to $\vec{\text{BA}}\dots(\text{iv})$
Its direction is parallel to the axis of dipole from positive to negative charge. It may be noted clearly from equations (ii) and (iv) that electric field strength due to a short dipole at any point is inversely proportional to the cube of its distance from the dipole and the electric field strength at axial position is twice that at broad-side on position for the same distance.
Important: Note the important point that the electric field due to a dipole at large distances falls off as $\frac{1}{\text{r}^3}$ and not as $\frac{1}{\text{r}^2}$ as in the case of a point charge.

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